Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in JAVA):

Solution:

import java.util.*;
import java.io.*;

public class Shades {

    static int N, M, K;
    static String s;
    static StringTokenizer st;
    static int[] d;

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        st = new StringTokenizer(br.readLine());
        
        int sy = Integer.parseInt(st.nextToken());
        int a = Integer.parseInt(st.nextToken());
        int b = Integer.parseInt(st.nextToken());

        pair A = new pair(a, sy);
        pair B = new pair(b, sy);

        int N = Integer.parseInt(br.readLine());

        pair[] fences = new pair[N];
        int[] len = new int[N + 1];
        for (int i = 0; i < N; i++) {
            st = new StringTokenizer(br.readLine());
            int s = Integer.parseInt(st.nextToken());
            int e = Integer.parseInt(st.nextToken());

            len[i + 1] = len[i] + (e - s);
            fences[i] = new pair(s, e);
        }

        Arrays.sort(fences);

        int Q = Integer.parseInt(br.readLine());
        for (int i = 0; i < Q; i++) {
            //System.out.println("");
            st = new StringTokenizer(br.readLine());
            int x = Integer.parseInt(st.nextToken());
            int y = Integer.parseInt(st.nextToken());

            pair p = new pair(x, y);

            double start = xintercept(A, p);
            double end = xintercept(B, p);

            double mult = ((double) y + Math.abs(sy)) / y;

            double tot = 0;
            int bot = 0;
            int top = N - 1;
            int w1 = 0;

            while (bot <= top) {
                int mid = (bot + top) >> 1;
                if (fences[mid].a <= start) {
                    w1 = Math.max(w1, mid);
                    bot = mid + 1;
                } else {
                    top = mid - 1;
                }
            }

            boolean done = false;
            if (fences[w1].b > start) {
                done = true;
                tot += Math.max(0,Math.min(end, fences[w1].b) - Math.max(start,fences[w1].a));
            }

            bot = 0;
            top = N - 1;
            int w2 = N-1;

            while (bot <= top) {
                int mid = (bot + top) >> 1;
                if (fences[mid].b >= end) {
                    w2 = Math.min(w2, mid);
                    top = mid - 1;
                } else {
                    bot = mid + 1;
                }
            }

            
            if (fences[w2].a < end && (w1 != w2 || !done)) {
                tot += Math.max(0,Math.min(end,fences[w2].b) - Math.max(start,fences[w2].a));
            }

            
            if (w2 > w1) {
                tot += len[w2] - len[w1+1];
            }
            //System.out.println((i+1)+ " : " +  w1 + " " + w2 + " " + start + " " + end + " " + mult + " " + tot + " " + (tot * mult));
            //out.print((i+1) + " ");
            out.printf("%.7f \n", (tot * mult));

        }
        
        //System.out.println(Arrays.toString(len));

        out.close();
    }

    static class pair implements Comparable<pair> {

        int a, b;

        public pair(int _a, int _b) {
            this.a = _a;
            this.b = _b;
        }

        @Override
        public int compareTo(pair t) {
            return (a == t.a) ? b - t.b : a - t.a;
        }
    }

    public static double xintercept(pair a, pair b) {
        //System.out.println(a.a + " " + a.b + " --- " + b.a + " " + b.b);
        if (a.a == b.a) {
            return a.a;
        }
        double m = (double) (a.b - b.b) / (a.a - b.a);
        //System.out.println("slope =" + m);
        if (m == 0) {
            return m;
        }
        return a.a - a.b / m;
    }
}